Divisibility Induction $17|18^{(5n+1)}+13^{(4n+1)}+3$ Prove that by induction
$17|18^{(5n+1)}+13^{(4n+1)}+3$ for all n∈N.
So far I'm stuck on the proof for n=k+1:
When n=k+1:
$$RHS=18^{5k+6}+13^{4k+5}+3$$
$$= (18^{5k+1}+13^{4k+1}+3)+[(18^5-1)18^{5k+1}+(13^4-1)13^{4k+1}]$$
From the assumption n=k, I can prove the first part is divisible by 17, but unsure of how to prove for the second part.
Any help is appreciated.
Edit:word
 A: $17 | (18^5-1)$ and $17 | (13^4-1)$, you can get this by computing these expressions. And after that you have three terms - each divisible by 17 so their sum is alos divisible by 17.
A: Hint:
$18^5-1=(18-1)( m)=17 m$
$13^4-1=(13^2+1)(13^2-1)=170(168)$
A: Hint:
If $f(n)=18^{5n+1}+13^{4n+1}+3,$
$f(m+1)-13^4f(m)=18^{5(m+1)+1}-13^418^{5m+1}-3(13^4-1)$
$=((18^5-1)-(13^4-1)18^{5m+1}-3(13^4-1)$
So, it is sufficient to establish $17$ divides $13^4-1,18^5-1$
A: For completeness, here is another way to achieve the same result:
$$RHS=18^{5k+6}+13^{4k+5}+3$$
$$=18^5 \cdot 18^{5k+1}+ 13^4 \cdot 13^{4k+1}+3$$
$$=13^4 \cdot 18^{5k+1}+ 13^4 \cdot 13^{4k+1}+ (18^5-13^4)\cdot 18^{5k+1}+3$$
$$=13^4 (18^{5k+1}+ 13^{4k+1})+ (18^5-13^4)\cdot 18^{5k+1}+3$$
$$=13^4 (18^{5k+1}+ 13^{4k+1})+3\cdot13^4+ (18^5-13^4)\cdot 18^{5k+1}+3-3 \cdot13^{4}$$
$$=13^4 (18^{5k+1}+ 13^{4k+1}+3)+ (18^5-13^4)\cdot 18^{5k+1}+3-3 \cdot13^{4}$$
The first term is divisible by $17$ by the induction hypothesis. The second term is divisible by $17$ because $17$ divides $18^5-13^4 = 1861007$, and the same goes for $3 - 3 \cdot 13^4 = -85680$.
A: Just get a calculator and do the math.
$$18^5-1 = 1889567 = 17 \times 111151$$
$$13^4 - 1 = 28560 = 17 \times 1680$$
